I could not integrate using MatLab, Can you please help me?

In solving a problem I need to integrate the following function with respect to 't' from the limit 0 to t.

3*2^(1/2)*(1 - cos(4*t))^(1/2)*(a^2 + c^2)^(1/2)

I used the following commands but got the same result as given herewith.

>> syms a c t real

mag_dr = 3*2^(1/2)*(1 - cos(4*t))^(1/2)*(a^2 + c^2)^(1/2)

>> int(mag_dr,t,0,t)

ans =

int(3*2^(1/2)*(1 - cos(4*t))^(1/2)*(a^2 + c^2)^(1/2), t, 0, t)

Let me know the best way(s) to tackle this type of problem.

[Walter Roberson]

Integration by parts, using a change of variables u=4*t

Answer 1

[Dyuman Joshi]

syms t a c
fun = 3*2^(1/2)*(1 - cos(4*t))^(1/2)*(a^2 + c^2)^(1/2);
z = int(fun,t); %gives indefinite integral
%result of integration, z = -(3*sin(4*t)*(a^2 + c^2)^(1/2))/(2*(sin(2*t)^2)^(1/2));
t=0;
res = z - subs(z); 
%obtain final result by evaluating the integral, z(t)-z(0), by assigning t & using subs()

However, you will not get the result. See @Walter Roberson's comment below for more details.

[Walter Roberson]

Not quite.

syms t a c
fun = 3*2^(1/2)*(1 - cos(4*t))^(1/2)*(a^2 + c^2)^(1/2);
z = int(fun,t); %gives indefinite integral
char(z)

ans = '-(3*sin(4*t)*(a^2 + c^2)^(1/2))/(2*(sin(2*t)^2)^(1/2))'

z0 = limit(z, t, 0, 'right');
char(z0)

ans = '-3*(a^2 + c^2)^(1/2)'

res = simplify(z - z0);
char(res)

ans = '3*(a^2 + c^2)^(1/2) - (3*sin(4*t)*(a^2 + c^2)^(1/2))/(2*(sin(2*t)^2)^(1/2))'

fplot(subs(z, [a,c], [1 2]), [-5 5])

fplot((subs(fun,[a,c], [1 2])), [-5 5])

That is, the problem is that the integral is discontinuous at t = 0 and that is why int() cannot resolve it.

Answer 2

[Walter Roberson]

syms a c t real
mag_dr = 3*2^(1/2)*(1 - cos(4*t))^(1/2)*(a^2 + c^2)^(1/2)

mag_dr = 

z = int(mag_dr, t)

z = 

z - limit(z, t, 0, 'right')

ans = 

The integral is discontinuous at 0, which is why it cannot be resolved by MATLAB.

[Dyuman Joshi]

@Walter Roberson, is it not a good idea to evaluate symbolic integration limit using subs() as I did as well as in general?

[Walter Roberson]

Look more carefully at what you did. You coded

t=0;
res = z - subs(fun); 

You are substituting t = 0 into fun not into z .

If you substitute t = 0 into z then you will get an error about division by 0.

You are doing the equivalent of trying to evaluate int(f, x, a, b) as F(b) - f(a) instead of F(b) - F(a)

Example:

Integral of sin(x) for a full rotation, 2*pi, is 0 because the area above the axes is exactly balanced by the area below the axes. So int(sin(x), x, 0, 2*pi) should be 0. But your method would use subs(int(sin(x),x), x, 2*pi) - subs(sin(x),x,0) which would be subs(cos(x), x, 2*pi) - sin(0) which would be cos(2*pi) - 0 which would be 1 - 0 which is 1 which is incorrect. What is needed instead is subs(int(sin(x),x),x,2*pi) - subs(int(sin(x),x),x,0) which would be cos(2*pi) - cos(0) which would be 1 - 1 which would be 0 which is the correct solution.

[Dyuman Joshi]

The wrong substitution was a mistake on my part, mostly cause I did it in a hurry. I have edited my nswer accordingly as well. Other than that, is subs() a good approach or would you recommend otherwise?

[Walter Roberson]

limit() is more robust than subs() for cases like this. But limit() is sometimes quite expensive to calculate, or is beyond MATLAB's ability to calculate, even in some finite cases.

Answer 3

[Sindhu Karri]

Hii

The "int" function cannot solve all integrals since symbolic integration is such a complicated task. It is also possible that no analytic or elementary closed-form solution exists.

For definite integrals, a numeric approximation can be performed by using the "integral" function.

[Walter Roberson]

It does exist.